At first we note that for any \(a/q\)  with \((a,q)=1\)  we have \(|G(a/q)|\le 1\) . Hence, by Corollary REF  with \(\varepsilon =\sigma (k)/10\)  and the fact that \(d\ge 2^k+1\)  we obtain the estimate
\(\Big |\sum _{\genfrac{}{}{0.0pt}{}{a=1}{(a, q)=1}}^qG(a/q)^de\left( -Na/q \right)\Big |\le \sum _{\genfrac{}{}{0.0pt}{}{a=1}{(a, q)=1}}^q|G(a/q)|^{2^k+1}\lesssim _k q^{1-(2^{k}+1)9\sigma (k)/10}\lesssim q^{-11/10}.\) 
Consequently, for \(d\ge 2^k+1\) , we get
\(|\mathfrak {S}(N)|\le \sum _{q=1}^\infty \Big |\sum _{\genfrac{}{}{0.0pt}{}{a=1}{(a, q)=1}}^qG(a/q)^de\left( -Na/q \right)\Big |\lesssim _k\sum _{q=1}^\infty q^{-11/10}\lesssim _k 1.\) 
Since \(\mathfrak {S}(N)>0\)  we obtain the upper bound for \(\mathfrak {S}(N)\) . To prove the lower bound let us define
\(\chi _N(p):=\sum _{h=0}^\infty \sum _{\genfrac{}{}{0.0pt}{}{a=1}{(a, p^h)=1}}^{p^h}G(a/p^h)^de\left( -Na/p^h \right),\quad p\in \mathbb {P}.\) 
It is known, see [1]}, that
\(\mathfrak {G}(N)=\prod _{p\in \mathbb {P}}\chi _N(p).\) 
Moreover, by [1]} there is a natural number \(\gamma _k\)  which depends only on \(k\in \mathbb {N}\)  such that
\(\chi _N(p)\ge p^{\gamma _k(1-d)}>0,\) 
provided that \(d\ge 2^{k}+1\) . By following the proof of [1]} and using the estimate (REF ) one can prove that there is \(p_0=p_0(k)\)  such that
\(1/2\le \prod _{p\ge p_0}\chi _N(p)\le 3/2.\) 
Hence, by (REF ) and (REF ) we can estimate
\(\mathfrak {G}(N)\ge 1/2\prod _{p\le p_0}\chi _N(p)\ge 1/2\prod _{p\le p_0}p^{\gamma _k(1-d)}\ge 1/2^d\prod _{p\le p_0}p^{-\gamma _kd}=:a_k^d>0.\) 
